Closed Interval Defined by Absolute Value

Theorem

Let $\xi, \delta \in \R$ be real numbers.

Let $\delta > 0$.

Then:

$\set {x \in \R: \size {\xi - x} \le \delta} = \closedint {\xi - \delta} {\xi + \delta}$

where $\closedint {\xi - \delta} {\xi + \delta}$ is the closed real interval between $\xi - \delta$ and $\xi + \delta$.

$\ds \size {\xi - x}$

$\le$

$\ds \delta$

$\ds \leadstoandfrom \ \ $

$\ds -\delta$

$\le$

$\ds \xi - x \le \delta$

$\ds \leadstoandfrom \ \ $

$\ds \delta$

$\ge$

$\ds x - \xi \ge -\delta$

$\ds \leadstoandfrom \ \ $

$\ds \xi + \delta$

$\ge$

$\ds x \ge \xi - \delta$

$\ds \leadstoandfrom \ \ $

$\ds \xi - \delta$

$\le$

$\ds x \le \xi + \delta$

rearranging

By definition of closed real interval:

$\closedint {\xi - \delta} {\xi + \delta} = \set {x \in \R: \xi - \delta \le x \le \xi + \delta}$

Hence the result.

$\blacksquare$

$\set {x \in \R: \size {x - \xi} \le \delta} = \closedint {\xi - \delta} {\xi + \delta}$

which is immediate from:

$\size {x - \xi} = \size {\xi - x}$